A chain of uniform mass `m` and length `L` is held on a frictionless table in such a way that its `(1)/(n)`th part is hanging below the edge of table. The work done to pull the hanging part of chain is `:-`

Options

A. `(mgL^(2))/(2n^(2))`
B. Zero
C. `(mgL)/(2n)`
D. `(mgL)/(2n^(2))`

Answer

D

Explanation

Required work done`=` change in potential energy of chain.
Now let Potential energy `(U)=0` at table level so potential energies of chain
`U_(i)=-mg((L)/(2n))=-((M)/(L))(L)/(n)g((L)/(2n))=-(MgL)/(2n^(2))`,
`U_(f)=0`
Work done `=U_(f)-U_(i)=0-(-(MgL)/(2n^(2)))=(MgL)/(2n^(2))`